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Maths M.C.Q (1 Marks)

Model Paper 2 · CBSE STD 11 Science (CBSE - English Medium). 18 questions.

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M.C.Q (1 Marks)

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18 questions · timed · auto-graded

MCQ 11 Mark
In an examination, a candidate has to pass in each of the five subjects. In how many ways can he fail?
  • ✓
    $31$
  • B
    $10$
  • C
    $21$
  • D
    $5$
Answer
Correct option: A.
$31$
The candidate can fail by failing in $1$ or $2$ or $3$ or $4$ or $5$ subjects out of $5$ in each case.
$\therefore$ required number of ways $={ }^5 C _1+{ }^5 C _2+{ }^5 C _3+{ }^5 C _4+{ }^5 C _5$
$={ }^5 C _1+{ }^5 C _2+{ }^5 C _{(5-3)}+{ }^5 C _{(5-4)}+1$
$={ }^5 C _1+{ }^5 C _2+{ }^5 C _2+{ }^5 C _1+1$
$=2\left({ }^5 C _1+{ }^5 C _2\right)+1$
$=2\left(5+\frac{5 \times 4}{2 \times 1}\right)+1$
$=(30+1)$
$=31$
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MCQ 21 Mark
$\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}$ is equal to:
  • ✓
    $n a^{ n -1}$
  • B
    $1$
  • C
    $na^n$
  • D
    $na$
Answer
Correct option: A.
$n a^{ n -1}$
$\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}$
$=\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}[\because f(x)$ exists, $\lim _{x \rightarrow a} f(x)$
$=\lim _{x \rightarrow a^{+}} f(x)]$
$=\lim _{h \rightarrow 0} \frac{(a+h)^n-a^n}{a+h-a}$
$=\lim _{h \rightarrow 0} a^n \frac{\left[\left(1+\frac{A}{a}\right)^n-1\right]}{h}$
$=a^{ n } \lim _{h \rightarrow 0}\left[1+ n \cdot \frac{h}{a}+\frac{n(n-1)}{2!} \frac{h^2}{a^2} \ldots+\ldots-1\right]$
$=a^{ n } \lim _{h \rightarrow 0}\left[\frac{n}{a}+\frac{h(h-1)}{2!} \frac{h}{a^2}+\ldots\right]$
$=a^{ n } \frac{n}{a}$
$=n a^{ n -1}$
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MCQ 31 Mark
$\sin 18^{\circ}=?$
  • A
    $\frac{(\sqrt{3}+1)}{2}$
  • B
    $\frac{(\sqrt{3}-1)}{2}$
  • C
    $\frac{(\sqrt{5}+1)}{4}$
  • D
    $\frac{(\sqrt{5}-1)}{4}$
Answer
(d) $\frac{(\sqrt{5}-1)}{4}$
Explanation: Remember $\sin 18^{\circ}=\frac{(\sqrt{5}-1)}{4}$
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MCQ 51 Mark
Solve the system of inequalities $(x+5)-7(x-2) \geq 4 x+9,2(x-3)-7(x+5) \leq 3 x-9$
  • ✓
    $\frac{-9}{4} \leq x \leq 1$
  • B
    $-4 \leq  x  \leq 1$
  • C
    $-1 \leq x \leq 1$
  • D
    $-4 \leq x \leq 4$
Answer
Correct option: A.
$\frac{-9}{4} \leq x \leq 1$
$(x+5)-7(x-2) \geq 4 x+9$
$\Rightarrow x+5-7 x+14 \geq 4 x+9$
$\Rightarrow-6 x+19 \geq 4 x+9$
$\Rightarrow-6 x-4 x \geq 9-19$
$\Rightarrow-10 x \geq-10$
$\Rightarrow x \leq 1$
$\Rightarrow x \in(-\infty, 1]$
$2(x-3)-7(x+5) \leq 3 x-9$
$\Rightarrow 2 x-6-7 x-35 \leq 3 x<9$
$\Rightarrow-5 x-41 \leq 3 x-9$
$\Rightarrow-5 x-3 x \leq 41-9$
$\Rightarrow-8 x \leq 32$
$\Rightarrow-x \leq \frac{32}{8}=4$
$\Rightarrow x \geq-4$
$\Rightarrow x \epsilon[-4, \infty)$
Hence the solution set is $[-4, \infty) \cap(-\infty, 1]=[-4,1]$
Which means $-4 \leq x \leq 1$
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MCQ 61 Mark
If $x=99^{50}+100^{50}$ and $y=(101)^{50}$ then
  • ✓
    $x < y$
  • B
    $x > y$
  • C
    $x = y$
  • D
    $x \geq y$
Answer
Correct option: A.
$x < y$
Given $x=99^{50}+100^{50}$ and $y=(101)^{50}$
Now $y, =(101)^{50}=(100+1)^{50}={ }^{50} C_0(100)^{50}+{ }^{50} C_1(100)^{49}+{ }^{50} C_2(100)^{48}+\ldots .+{ }^{50} C_{50} \ldots .$
$(i)$ Also $(99)^{50}=(100-1)^{50}=={ }^{50} C_0(100)^{50}-{ }^{50} C_1(100)^{49}+{ }^{50} C_2(100)^{48}-\ldots .+{ }^{50} C_{50} \ldots$
$(ii)$ Now subtract equation (ii) from equation $(i),$
we get $(101)^{50}-(99)^{50}=2\left[{ }^{50} C_1 \quad(100)^{49}+{ }^{50} C_3 \quad(100)^{47}+\ldots\right]$
$=2\left[50(100)^{49}+\frac{50 \times 49 \times 48}{3 \times 2 \times 1}(100)^{47}+\ldots\right]$
$=(100)^{50}+2\left(\frac{50 \times 49 \times 48}{3 \times 2 \times 1}(100)^{47}\right)$
$\Rightarrow(101)^{50}-(99)^{50}>(100)^{50}$
$\Rightarrow(101)^{50} > (100)^{50}+(99)^{50}$
$\Rightarrow y > x$
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MCQ 71 Mark
$(\sqrt{5}+1)^4+(\sqrt{5}-1)^4$ is
  • A
    an irrational number
  • B
    a negative real number
  • ✓
    a rational number
  • D
    a negative integer
Answer
Correct option: C.
a rational number
We have $(a+b)^n+(a-b)^n$
$=\left[{ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2}
b^2+{ }^n C_3 a^{n-3} b^3+\ldots . .+{ }^n C_n b^n\right]+$
${\left[{ }^n
C_0 a^n-{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2-{ }^n C_3 a^{n-3}
b^3+\ldots . .+(-1)^n \cdot{ }^n C_n b^n\right]}$
$=2\left[{ }^n C_0 \quad
a^n+{ }^n C_2 a^{n-2} b^2+\ldots\right]$
Let $a =\sqrt{5}$ and $b =1$ and $n =4$
Now we get $(\sqrt{5}+1)^4+(\sqrt{5}-1)^4$
$=2\left[{ }^4 C_0(\sqrt{5})^4+{ }^4C_2(\sqrt{5})^2 1^2+{ }^4 C_4(\sqrt{5})^0 1^4\right]$
$=2[25+30+1]=112$
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MCQ 81 Mark
Let $S=\{x \mid x$ is a positive multiple of $3$ less than $100\}$
$P=\{x \mid x$ is a prime number less than $20\}$. Then $n(S)+n(P)$ is
  • ✓
    $41$
  • B
    $30$
  • C
    $34$
  • D
    $33$
Answer
Correct option: A.
$41$
We have to find ,$n(S) + n(P)$
$S = (x | x$ is a positive multiple of $3$ less than $100)$
$\Rightarrow S=\{3,6,9,12,15, \ldots \ldots, 99\}$
$\Rightarrow n(S)=\frac{99}{3}=33$
$P=\{x \mid x$ is a prime number less than $20\}$
$\Rightarrow P=\{2,3,5\ aq\ 7,11,13,17,19\}$
$\Rightarrow n(P)=8$
$n(S)+n(P)=33+8=41$
Therefore, answer is $41$
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MCQ 91 Mark
If $A - B =\frac{\pi}{4}$, then $(1+\tan A )(1-\tan B )$ is equal to
  • ✓
    $2$
  • B
    $0$
  • C
    $1$
  • D
    $3$
Answer
Correct option: A.
$2$
$\tan (A-B)=\tan \frac{\pi}{4}$
$\Rightarrow \frac{\tan A-\tan B}{1+\tan A \tan B}=1$
$\Rightarrow \tan A=\tan B=1+\tan A \tan B \ldots (i)$
Now,
$=(1+\tan A)(1-\tan B)=1+\tan A=\tan B=\tan A \tan B$
$=1+1+\tan A \tan B=\tan A \tan B \text { (Using eq. (i)) }$
$=2$
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MCQ 111 Mark
The number of ways in which 5 + and 5 – signs can be arranged in a line such that no two – signs occur together is
  • A
    P(5, 5)
  • B
    C(5, 5)
  • C
    P(6, 5)
  • D
    C(6, 5)
Answer
(d) C(6, 5)
Explanation: Since all the plus signs are identical, we have number of ways in which 5 plus signs can be arranged = 1.
Now we will have 6 empty slots between these 5 identical + signs
Hence the number of possible places of - sign = 6
Therefore the number of ways in which the 5 minus sign can take any of the possible 6 places = C(6, 5)  
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MCQ 121 Mark
If $z =(3+\sqrt{2} i)$ then $z \times z =$ ?
  • A
    11
  • B
    7
  • C
    $\sqrt{11}$
  • D
    5
Answer
(a) 11
Explanation: $zz =| z |^2=\left\{3^2+(\sqrt{2})^2\right\}=(9+2)=11$
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MCQ 131 Mark
The reflection of the point $(\alpha, \beta, \gamma)$ in the xy- plane is
  • A
    $(\alpha, \beta,-\gamma)$
  • B
    $(0,0, \gamma)$
  • C
    $(\alpha, \beta, 0)$
  • D
    $(-\alpha,-\beta, \gamma)$
Answer
(a) $(\alpha, \beta,-\gamma)$
Explanation: In xy-plane, the reflection of the point $(\alpha, \beta, \gamma)$ is $(\alpha, \beta,-\gamma)$
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MCQ 141 Mark
The lines $8 x+4 y=1,8 x+4 y=5,4 x+8 y=3,4 x+8 y=7$ form a
  • A
    Trapezium
  • B
    Rhombus
  • C
    Rectangle
  • D
    Square
Answer
(b) Rhombus
Explanation: On solving the equations $8 x+4 y=1$ and $4 x+8 y=3$, we get the point of intersection as $\left(\frac{-1}{2}, \frac{5}{12}\right)$
On solving the equations $8 x+4 y=5$ and $4 x+8 y=7$, we get the point of intersection as $\left(\frac{1}{4}, \frac{3}{4}\right)$
On solving the equations $8 x+4 y=1$ and $4 x+8 y=7$, we get the point of intersection as $\left(\frac{-5}{12}, \frac{13}{12}\right)$
On solving the equations $8 x+4 y=5$ and $4 x+8 y=3$, we the point of intersection as $\left(\frac{7}{12}, \frac{1}{12}\right)$
Let the points $A \left(\left(\frac{-1}{2}, \frac{5}{12}\right), B \left(\frac{7}{12}, \frac{1}{12}\right) C \left(\frac{1}{4}, \frac{3}{4}\right)\right.$ and $D \left(\frac{-5}{12}, \frac{13}{12}\right)$ be the vertices of the quadrilateral
Since the slopes of the opposite sides are equal the quadrilateral is a parallelogram
The slope of the diagonal AC is $\frac{5 / 12-3 / 4}{-1 / 2-1 / 4}=1$
The slope of the diagonal BD is $\frac{1 / 2-13 / 12}{7 / 12-(-5 / 12)}=-1$
Since the product of the slopes is -1 , the diagonals are perpendicular to each other.
Hence the parallelogram is a rhombus.
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MCQ 151 Mark
$\lim _{x \rightarrow 0} \frac{\sin x^n}{(\sin x)^m}, n > m >0$ is equal to
  • A
    $\frac{m}{n}$
  • ✓
    $0$
  • C
    $1$
  • D
    $\frac{n}{m}$
Answer
Correct option: B.
$0$
$ \lim _{x \rightarrow 0} \frac{\sin x^n}{(\sin x)^m} \cdot \frac{x^{m+n}}{x^{m+n}}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin x^n}{x^n} \cdot \frac{x \rightarrow 0}{(\sin x)^m} \cdot \frac{x^n}{x^m}$
$\Rightarrow 1.1^m \cdot x^{n-m}$
$\Rightarrow 1(0)=0$
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MCQ 161 Mark
The mean and S.D. of 1, 2, 3, 4, 5, 6 is
  • A
    $3, \frac{35}{12}$
  • B
    3,3
  • C
    $\frac{7}{2}, \sqrt{\frac{35}{12}}$
  • D
    $\frac{7}{2}, \sqrt{3}$
Answer
(c) $\frac{7}{2}, \sqrt{\frac{35}{12}}$
Explanation: Mean $=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=\frac{7}{2}$
S.D $=\sqrt{\frac{n^2-1}{12}}=\sqrt{\frac{36-1}{12}}=\sqrt{\frac{35}{12}}$
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MCQ 171 Mark
if $f \left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$ then $f ( x )=$ ?
  • ✓
    $\left(x^2-2\right)$
  • B
    $\left(x^2+1\right)$
  • C
    $\left(x^2-1\right)$
  • D
    $x^2$
Answer
Correct option: A.
$\left(x^2-2\right)$
$f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}$
$=\left(x+\frac{1}{x}\right)^2-2$
$\text { Put, }\left(x+\frac{1}{x}\right)=t$
$\Rightarrow f(t)=t^2-2$
$\therefore f(x)=x^2-2$
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MCQ 181 Mark
$\cos 40^{\circ}+\cos 80^{\circ}+\cos 160^{\circ}+\cos 240^{\circ}=$
  • A
    $\frac{1}{2}$
  • ✓
    $\frac{-1}{2}$
  • C
    $1$
  • D
    $0$
Answer
Correct option: B.
$\frac{-1}{2}$
$\cos 40^{\circ}+\cos 80^{\circ}+\cos 160^{\circ}+\cos 240^{\circ}$
$=2 \cos \left(\frac{40^{\circ}+80^{\circ}}{2}\right) \cos \left(\frac{40^{\circ}-80^{\circ}}{2}\right)+\cos 160^{\circ}+\cos \left(180^{\circ}+60^{\circ}\right)\left[\because \cos A +\cos B =2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$
$=2 \cos 60^{\circ} \cos \left(-20^{\circ}\right)+\cos 160^{\circ}-\frac{1}{2}$
$=2 \times \frac{1}{2} \cos 20^{\circ}+\cos 160^{\circ}=\frac{1}{2}$
$=\cos \left(180^{\circ}-20^{\circ}\right)+\cos 20^{\circ}-\frac{1}{2}$
$=-\cos 20^{\circ}+\cos 20^{\circ}-\frac{1}{2}$
$=-\frac{1}{2}$
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M.C.Q (1 Marks) - Maths STD 11 Science Questions - Vidyadip